Textbook: Vollhardt 6th Ed. (2010)

Chapter 22: Chemistry of Benzene Substituents: Alkylbenzenes, Phenols, and Benzamines

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Individual Problems

Individual Problems

Problem # 580

Draw all resonance forms for each species.

For the anion and cation species, used curved arrows. For the radical species, use hooks.

Solution

(This is similar to problem 569, which involved the allylic position.)

Notice that arrows always go from high electron density to low electron density. So for negatively charged species, the arrow starts at the lone pair. For positively charged species, the arrow starts at the double bond and goes towards the positive charge.

(Radical resonance uses hooks instead of arrows and so is a little different.)

Problem Details

MendelSet practice problem # 580 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Lewis Structures, Formal Charges, and Resonance, Benzene Side Chain Reactions and the Benzylic Position

Question Type: Concept (explain why this is so..)

Keyword(s): resonance, benzylic position

Problem # 755

The nitrosyl cation is shown below. Also shown are several proposed resonance arrows, only one of which is correct.

Draw the resonance forms that would follow from each set of arrows, and include formal charges. Which one is the correct resonance form? Explain your reasoning.

Solution

When evaluating possible Lewis structures the most important rule to follow is the octet rule: every atom must be surrounded by 8 and only 8 electrons (atoms in period 3 or below on the periodic table have d orbitals and so can contain more than 8 electrons, but that exception does not apply here).

Do any of the resonance forms drawn have atoms that break the octet rule? Yes- they all do! Except for resonance arrows in D. So D is the correct resonance form.

Problem Details

MendelSet practice problem # 755 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Lewis Structures, Formal Charges, and Resonance, Diazonium Chemistry (NaNO2, Sandmeyer, Nitrosyl)

Question Type: Concept (explain why this is so..)

Keyword(s): resonance, Lewis structures

Problem # 611

Draw in the arrows to show the electron flow and resonance forms in the nucleophilic aromatic substitution reaction below.

Note: Depending on the textbook, nucleophilic aromatic substitution is referred to as NAS, S_NAr, or addition-elimination.

Solution

SNAr is sort of like SN2, except the leaving group doesn't leave right away; a tetrahedral intermediate is formed first.

The trick with SNAr (or NAS, addition-elimination, etc.) is to draw resonance forms that stabilizes the negative charge that forms. That's why EWG increase the rate of S_NAr (they stabilize negative charges).

Problem Details

MendelSet practice problem # 611 submitted by Matt on July 10, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Aromatic Substitution and Benzyne

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): SNAr

Problem # 612

Draw a mechanism for the nucleophilic aromatic substitution (S_NAr) reaction below. Show all resonance forms of the intermediate.

Solution

For S_NAr to work you need to have electron withdrawing groups (EWG) either ortho or para to the leaving group.

This molecule has two leaving groups (chlorines), but only one chlorine has EWG ortho/para to it. So that's the carbon where the nucleophile (NH₃) will attack.

Why do EWG need to be ortho/para? Because that's the only way for the negative charge to be stabilized by resonance. Try it- if you have the NH₃ attack the other carbon with a chlorine, you will not be able to draw a resonance form that places the negative charge on one of the EWG.

Problem Details

MendelSet practice problem # 612 submitted by Matt on July 10, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Aromatic Substitution and Benzyne

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): SNAr

Problem # 615

Let's go through a benzyne reaction (also called elimination-addition).

In the reaction below, the strong base (NaNH₂) will form a benzyne intermediate, which when forms either ortho nitroaniline or meta nitroaniline.

Used curved arrows to show the formation of each intermediate and the final products.

Solution

First the ^-NH₂ acts as a base to eliminate a leaving group and form Benzyne. Then the ^-NH₂ acts as a nucleophile and attacks the benzyne, and the "leaving group" is the triple bond.

Problem Details

MendelSet practice problem # 615 submitted by Matt on July 10, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Aromatic Substitution and Benzyne

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): benzyne

Problem # 305

Rank each group of acids in order of decreasing acidity. (1 = most acidic)

Explain your reasoning. You will have to use more than one rule in your explanation (resonance, electronegativity, atomic radius, etc.).

Solution

Phenol is more acidic than cyclohexanol because the conjugate base of phenol (phenolate) has resonance while the conjugate base of cyclohexanol does not.

Thiophenol is more acidic than phenol because sulfur is larger than oxygen, and so RS^- is more stable than RO^-.

Problem Details

MendelSet practice problem # 305 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Acid Strength Trends and Conjugate Base Stability, Phenols (Claisen Rearrangement, Kolbe-Schmitt, Phenol EAS)

Question Type: Concept (explain why this is so..)

Keyword(s): resonance, acid strength

Problem # 581

Draw all products for the two reactions below.

The allylic alkene gives two products- the 1,2 product, and the 1,4 product.

However, the benzylic alkene only gives the one product (analogous to the 1,2 product), instead of multiple products (like the 1,4 product, 1,6 product, and 1,8 product). Why is this the case?

Solution

The allylic carbocation has two resonance forms, and so two positions where a nucleophile can attack, yielding two different products (1,2 and 1,4).

The benzylic carbocation has four resonance forms, and so in theory, four positions where a nucleophile can attack. But all but one of these resonance forms yield a product that is not aromatic, and so do not form.

Aromatic compounds are very stable; aromatic starting materials tend to only form products that are also aromatic.

Problem Details

MendelSet practice problem # 581 submitted by Matt on July 9, 2011.

Subject: Organic Chemistry

Subject Topic(s): Benzene Side Chain Reactions and the Benzylic Position

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): conjugate addition, allylic position, benzylic position

Problem # 1281

Allylic and benzylic halides tend to undergo both S_N1 and S_N2 substitution reactions at a faster rate than their alkyl counterparts.

For example, both allyl chloride and benzyl chloride undergo S_N2 reaction at a faster rate than propyl chloride.

The same holds true for SN1 reactions: a 2° allyl or benzyl halide undergoes S_N1 reaction faster than a 2° alkyl halide. Explain.

Solution

There are separate explanations for SN1 and SN2.

S_N1: The S_N1 mechanism involves a carbocation intermediate, and both allylic and benzylic carbocations have resonance, which increases the stability of their carbocations, and speeds up the rate of S_N1 reaction.

S_N2: This explanation is less obvious, and is probably only mentioned in passing in your orgo textbook, if at all.

The pi systems present in allylic and benzylic halides are able to overlap with the pi orbitals of the nucleophile and the leaving group in the 5-coordinate transition state of an S_N2 reaction. This orbital overlap lowers the energy of the transition state (which stabilizes it), and so increases the rate of S_N2 reaction.

There are other possible explanations as well (Which also vary by textbook). For one, an allylic halide is less sterically hindered than an alkyl halide (less hydrogens sticking out in 3D).

Also, the carbons on the double bond of allyl and benzyl compounds are sp² hybridized, and so are more electronegative than sp³ carbons. So the sp²carbons pull alway some electron density from the alpha carbon (the carbon attached to the leaving group), making it more electrophilic, and thus increasing S_N2 reactivity.

Problem Details

MendelSet practice problem # 1281 submitted by Matt on October 1, 2011.

Subject: Organic Chemistry

Subject Topic(s): Conjugated Dienes and the Allylic Position (Kinetic vs Thermodynamic Control), Benzene Side Chain Reactions and the Benzylic Position

Question Type: Concept (explain why this is so..)

Keyword(s): E1 mechanism, SN2 mechanism, allylic position, benzylic position

Problem # 616

Let's go through another way to make benzyne.

First, let's form a Grignard reagent. Then, let's elminate to form benzyne.

Solution

Halogens get more reactive towards Grignard formation as we go down the periodic table, so Mg⁰ will form a Grignard with Br before F.

Grignards behave like negatively charged carbon atoms- very unstable. So even though fluorine is a poor leaving group, the negative carbon will cause it to eliminate to form benzyne.